GO Classes Set Theory And Algebra Practice Set 1 | Question: 15

Question

Let $R \subseteq \mathbb{N} \times \mathbb{N}$ be a relation ( $a$ binary relation) on the set of natural numbers defined as follows:
$$
(x, y) \in R \Leftrightarrow x+y \geq 18 \text {. }
$$

• A. is $R$ reflexive? Prove your answer.
• B. is R symmetric? Prove your answer.
• C. Is $R$ antisymmetric? Prove your answer.
• D. Is $\mathrm{R}$ transitive? Prove your answer.

Answer 1

A. To determine if R is reflexive, we need to check if for every element x ∈ N, (x, x) ∈ R. In this case, (x, x) ∈ R if and only if x + x ≥ 18, which simplifies to 2x ≥ 18 or x ≥ 9.

 

Since x is a natural number, it can take on any value greater than or equal to 9. Therefore, for every x ≥ 9, (x, x) ∈ R, making R reflexive.

 

B. To determine if R is symmetric, we need to check if for every pair (x, y) ∈ R, (y, x) ∈ R. Let's consider an example to analyze this.

 

Suppose we have (x, y) ∈ R, which means x + y ≥ 18. Now, if we swap the positions of x and y, we get (y, x). We need to check if (y, x) also satisfies the condition x + y ≥ 18.

 

When we swap x and y, the inequality becomes y + x ≥ 18, which is equivalent to x + y ≥ 18. Since the inequality is symmetric, if (x, y) satisfies it, then (y, x) also satisfies it.

 

Hence, for every (x, y) ∈ R, (y, x) ∈ R, making R symmetric.

 

C. To determine if R is antisymmetric, we need to check if for every distinct pair (x, y) ∈ R and (y, x) ∈ R, it implies that x = y. Let's consider an example to analyze this.

 

Suppose we have (x, y) ∈ R and (y, x) ∈ R, which means x + y ≥ 18 and y + x ≥ 18. From these two inequalities, we can conclude that x + y = y + x = 18.

 

Since x + y = y + x, we can subtract y from both sides to get x = y.

 

Hence, for every distinct (x, y) ∈ R and (y, x) ∈ R, x = y, making R antisymmetric.

 

D. To determine if R is transitive, we need to check if for every (x, y) ∈ R and (y, z) ∈ R, it implies that (x, z) ∈ R. Let's consider an example to analyze this.

 

Suppose we have (x, y) ∈ R and (y, z) ∈ R, which means x + y ≥ 18 and y + z ≥ 18. We want to prove that (x, z) ∈ R, which means x + z ≥ 18.

 

Adding the inequalities x + y ≥ 18 and y + z ≥ 18, we get (x + y) + (y + z) ≥ 18 + 18, which simplifies to x + 2y + z ≥ 36.

 

Since x + y ≥ 18, we can replace it with 18 in the inequality above to get 18 + y + z ≥ 36. Subtracting 18 from both sides gives y + z ≥ 18.

 

Therefore, we have shown that if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R. Hence, R is transitive.

 

In summary:

A. R is reflexive.

B. R is symmetric.

C. R is antisymmetric.

D. R is transitive.